In the figure, O Is the centre of the circle and HP is parallel to KL. NR = NP, ∠OKH = 56° and ∠RPN = 55°. Find
- ∠LKS
- ∠KLN
(a)
OH = OK = Radius
∠HKO = ∠KHO = 56° (Isosceles triangle, OHB)
∠KOS
= 56° + 56°
= 112° (Exterior angle of a triangle)
OK = OS = Radius
∠OSK
= (180° - 112°) ÷ 2
= 68° ÷ 2
= 34°
∠LKS = 34° (Alternate angles, KL//HE)
(b)
NR = NP
∠NPR = ∠NRP = 55° (Isosceles triangle NPF)
∠PNR
= 180° - 55° - 55°
= 70°
∠SNL = ∠PNR = 70° (Vertically opposite angles)
∠KLN
= 180° - 70°
= 110° (Interior angles)
Answer(s): (a) 34°; (b) 110°